Compare mass of pure NaOH in each of the aqueous solution
50 g of 40% (w/w) NaOH
50 ml of 50% (w/v) NaOH ['d_(soln)=1.2 g/(mL)']

Options

A. (ii)>(i)
B. (i) >(ii)
C. (i) =(ii)
D. Mass in (i) is double the mass in (ii)

Answer

To compare the mass of pure NaOH in each of the given aqueous solutions, we will calculate the amount of NaOH in both solutions step by step.

### Step 1: Calculate the mass of pure NaOH in the first solution (50 g of 40% w/w NaOH)

1. **Understanding the percentage**: A 40% w/w NaOH solution means there are 40 grams of NaOH in 100 grams of the solution.
2. **Setting up the proportion**: Since we have 50 grams of this solution, we can find the amount of NaOH in it using the following proportion:
\[ \text{Mass of NaOH} = \left(\frac{40 \, \text{g NaOH}}{100 \, \text{g solution}}\right) \times 50 \, \text{g solution} \]
3. **Calculating**:
\[ \text{Mass of NaOH} = \frac{40}{100} \times 50 = 20 \, \text{g} \]

### Step 2: Calculate the mass of pure NaOH in the second solution (50 ml of 50% w/v NaOH)

1. **Understanding the percentage**: A 50% w/v NaOH solution means there are 50 grams of NaOH in 100 mL of the solution.
2. **Setting up the proportion**: We have 50 mL of this solution, so we can find the amount of NaOH in it using the following proportion:
\[ \text{Mass of NaOH} = \left(\frac{50 \, \text{g NaOH}}{100 \, \text{mL solution}}\right) \times 50 \, \text{mL solution} \]
3. **Calculating**:
\[ \text{Mass of NaOH} = \frac{50}{100} \times 50 = 25 \, \text{g} \]

### Step 3: Compare the masses of pure NaOH in both solutions

- From Step 1, we found that the mass of pure NaOH in the first solution is **20 g**.
- From Step 2, we found that the mass of pure NaOH in the second solution is **25 g**.

### Conclusion

Since 25 g (from the second solution) is greater than 20 g (from the first solution), we conclude that:

**The mass of pure NaOH in the second solution is greater than that in the first solution.**

### Final Answer

Mass of pure NaOH in the second solution is greater than mass of pure NaOH in the first solution.

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